application of first order ordinary differential equations

EMDADUL HAQUE MILONehmilon24171@gmail.com

mdehmilon24171@gmail.com

DEPERTMENT OF STATISTICS,UNIVERSITY OF RAJSHAHI..RAJSHAHI, BANGLADESH

6205&6206

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DEPARTMENT OF statistics, UNIVERSITY OF

RAJSHAHI.

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Welcome to all of our presentation of

GROUP-D

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1. Md. Jenarul Islam 1510224120

2. Md. Lukman Joni 1510924175

3. Md. Shamim Hossen 15010424107

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4. Md. Sharifuzzaman 1510224164

5. Md. Amdadul Hasan 1510924174

6. Md. Emdadul Haque (Milon) 1510924171

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7. Md. Abdul Bari 1510324161

8. Md. Naoaj Sharif 1510524171

9. Md. Shahinur alam 1510824130

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10. Md. Abdul Alim 1510824155

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TOPICS OF OUR PRESENTATION

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1. Applications of first order ordinary differential equation.2. Orthogonal trajectory. &3. Oblique trajectory.

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Applications of 1st order ordinary differential equation.

1.

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1st order ordinary differential equation:Definition of 1st order ordinary differential equation: 1st order ordinary differential equation is one kind of differential equation. A differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable and which has only one order derivatives, is called a 1st order ordinary differential equation.

Example: 1. is a 1st order ordinary differential equation. Here y is a dependent variable and x is a independent variable and is the derivative term which order is one, so it is a 1st order ordinary differential equation.

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Standard form of 1st order ordinary differential equation:

The standard form of 1st order ordinary differential equation is or the differential form In the form (1) it is clear from the notation itself that y is regarded as the dependent variable and x as the independent one.But in the form (2) we may actually regard either variable as the dependent one and the other as the independent.

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Applications of 1st order ordinary differential equation :

There are a lot of applications of 1st order ordinary differential equation in our real life in various sectors. Some of these are given below:

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1.Cooling/Warming Law (use in physics)2.Population Growth and Decay (in stat..)3.Radio-Active Decay and Carbon Dating 4.Mixture of Two Salt Solutions(in

chemistry) 5.Series Circuits (in physics)6.Survivability with AIDS (in medicine)7.Draining a tank (in engineering)

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8. Economics and Finance ( in economics)9. Mathematics Police Women10. Drug Distribution in Human Body ( in biology)11. A Pursuit Problem 12. Harvesting of Renewable Natural Resources (in agriculture)13.Determining the motion of a projectile, rocket,

satellite or planet (in engineering). 14.Determining the charge or current in a electric circuit(in physics)

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15. Determination of curves that have certain geometrical properties. 16. Conduction of heat in a rod or in a slab. 17. Determining the vibrations of a wire or membrane. And so on.

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Some applications of 1st order ordinary differential equation in engineering

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Falling stone

Parachute

Water level tank

Vibratingspring

Beats of vibrating system

Current circuit

Pendulum

Prey model

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Lets see some applications of 1st order ordinary differential equation with example.

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1. Population Growth and Decay (in statistics)Problem: A population grows at the rate of 5% per year. How long does it takes for the population to double?Solution: Let the initial population be p0 and let the population after t years be p.Then we get, ⇒which is a separable differential equation.Now we integrating (1) and we get,

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⇒ We put the initial value in (2) i.e.

⇒We get from (2) and (3), ⇒ ⇒

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when, p=2p0 then, ⇒ Hence the population is double in 14 years.

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2. Determination of curves that have certain geometrical properties.

Problem: The slope of the tangent at a point (x, y) on a curve is If the curve passes through the point (3,-4). Find the equation of the curve. Solution: We know the slope of a curve at point (x, y) is Thus by the problem Which is a separable differential equation.

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Now integrating (1) and we get, ⇒ where c is an arbitrary constant.Since the curve passes through the point (3, -4) thus from (2), ⇒

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Hence, , which is the required equation of the curve.

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3. Cooling/Warming Law (use in physics)

Example : When a chicken is removed from an oven, its temperature is measured at 3000F. Three minutes later its temperature is 200o F. How long will it take for the chicken to cool off to a room temperature of 70oF.

Solution: From the given problem ,

Tm = 70 and T=300 at for t=0. T(0)=300=70+c2e.0

This gives c2=230 For t=3, T(3)=200Now we put t=3, T(3)=200 and c2=230 then

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2301303αe

2313ln3

19018.02313ln

31

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Thus T(t)=70+230 e-0.19018t

We observe that furnishes no finite solution to T(t)=70 since limit T(t) =70.

t We observe that the limiting temperature is 700F.

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4. Carbon Dating

Example. Find the age of an object that has been excavated and found to have 90% of its original amount of radioactive Carbon 14.Solution: Using the equation y = y0ekt we see that we must find two things: (i) the value of k (ii) the value of t for which y0ekt = (90/100)y0, i.e., find t such that ekt = (9/10).

(i) Find the value of k: Rearranging the half-life equation and using the fact that the half-life is known to be 5570 years, we have

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−⇒ k =(ln2/half-life)⇒-k =ln2/5570 ≈ .0001244So k = −.0001244.

(ii) Find the value of t which makes ekt = 9/10:

⇒ e−.0001244t = 0.9 −⇒ .0001244t = ln(0.9) ⇒ t = [ln(0.9)/−.0001244] ≈ 878

Conclusion : Therefore the sample is approximately 878 years old.

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5.Radioactive Decay:

Radioactive decay: radioactive decay is the process of losing nuclei….Half-life: The Half-life of a radioactive element is the time required for half of the radioactive nuclei decay present in a sample (i.e. for the quantity to be reduced by one-half).

The half-life of an element is totally independent of the number of nuclei present initially, because this decay occurs exponentially, i.e. according to the differential equation dy/dt = ky, where k is some negative constant. (The value of the constant differs for the various radioactive elements.)• Let y0 be the number of radioactive nuclei present initially. Then the number

y of nuclei present at time t will be given by:

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⇒y = y0ekt

Since we are looking for the half-life, we wish to know the time t at which only 1/2(y0) nuclei remain:⇒Y0ekt =1/2(y0)

⇒ ekt =1/2

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⇒ kt = ln(1/2) = ln1−ln2 = −ln2 ⇒ t =ln2 −k

(Since k is negative, ln2 −k is positive.) Thus the half-life depends only on k.

The formula above is worth noting for future use:half-life = ln2 −k

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Example 13. The number of atoms of plutonium-210 remaining after t days, with an initial amount of y0 radioactive atoms, is given by: y = y0e(−4.95×10−3)t

Find the half-life of plutonium-210.

Solution: We see that for this element we have k = −4.95×10−3. Using the formula above we have:Half-life =(ln2/−k)={(ln2)/(4.95×10^−3)} ≈ 140

Therefore the half-life of plutonium-210 is approximately 140 days.

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2. Orthogonal Trajectory

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Orthogonal Trajectory:Definition: An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally, that is, at right angles (see Figure 7).

Figure:7

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Another Definition: Let

be a given one-parameter family of curves in the plane. A curve that intersects the curves of the family (2.1) at right angles is called an orthogonal trajectory of the given family.

Example: Consider the family of circles with center at the origin and radius c. Each straight line through the origin,

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is an orthogonal trajectory of the family of circles (2.2).Conversely, each circle of the family (2.2) is an orthogonal trajectory of the family of the straight lines (2.3). So the families (2.2) and (2.3) are orthogonal trajectories of each other.In figure (2.a.1) several members of the family of circles (2.2), drawn solidly and several members of the family of straight lines (2.3), drawn with dashes are shown.

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X

Y

Figure (2.a.1)

Member of the family of circles (2.2)

Member of the family of straight lines (2.3)

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Procedure for finding the orthogonal trajectories of a given family of curves:

Step 1. From the equation of the given family of curves, at first we find the differential equation of this family.Step 2. In the differential equation (2.5), so found in Step 1, replace f(x, y) by its negative reciprocal This gives the differential equation

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of the orthogonal trajectories.Step 3. At last we obtain a one parameter family or of solution of the differential equation (2.6), thus obtaining the desired family of orthogonal trajectories (except possibly for certain trajectories that are vertical lines and must be determined separately).Caution: In step 1, in finding the differential equation (2.5) of the given family, be sure to eliminate the parameter c during the process.

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Example of procedure of finding Orthogonal Trajectory:Example: Find the orthogonal trajectories of the family of parabolas Solution: Given the family of parabolas Step 1. We first find the differential equation of the given family (2.7)Differentiating (2.7), we obtain Eliminating the parameter c between equation (2.7) and (2.8)

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which is the differential equation of the family (2.7).Step 2. we now find the differential equation of the orthogonal trajectories by replacing in (2.9) by its negative reciprocal, obtaining Step 3. we now solve the differential equation (2.10). Here, Which is separable differential equation. Now integrating and we get,

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Where k is an arbitrary constant.This is the family of orthogonal trajectories of (2.7), it is clearly a family of ellipses with centers at the origin and major axes along the X axis.

Some members of the original family of parabolas and some of the orthogonal trajectories (ellipses) are shown in figure (2.a.2).

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X

Y

Member of the orthogonal

trajectories (ellipses) Member of the original family of parabolas

Figure (2.a.2)

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3. Oblique Trajectory

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Oblique trajectory:Definition: Let be a one parameter family of curves. A curve that intersects the curves of the family (3.1) at a constant angle is called an oblique trajectory of the given family.

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Procedure for finding the oblique trajectories of a given family of curves: Let be a family of curves.Suppose the slope of (3.2) is Let Ѳ be the angle of X axis and the tangent of the given family of curves. Thus,

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Ѳ 𝜑��

X

Y

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⇒Again let the angle of the tangent of oblique trajectories and X axis be .Now, by the theorem of triangle, we get ; where 𝛼 is the angle between two triangles. ⇒ ⇒

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Therefore, the slope of oblique trajectories is which is a differential equation.And after solving the equation, we have to get the function of oblique trajectories of the given family of curves.

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Example of procedure of finding oblique trajectory:Example: Find a family of oblique trajectories that intersects the family of straight lines at angle 45.Solution: Given the family of straight lines Differentiating, we obtain, Putting the value of c from (3.4) in (3.5) and we get,

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which is the differential equation of the given family of straight lines.Now, we replace in equation (3.6) by ; putting, Thus the differential equation of the desired oblique trajectories is Now, we solve the differential equation (3.7) and we get,

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which is the homogeneous differential equation. Now we write this in form we get, Let, Now from (3.8) and (3.9) we get,

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⇒

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⇒ ⇒ ⇒Which is the family of oblique trajectories.

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Reference: 1.Class lecture - Prof. Dr. Md. Golam Hossain, department of Statistics University of Rajshahi.2. Ross, S.L.(1989). Differential Equations, 4th ed., Wiley, N.Y.3. Google search: https://www.google.com.bd/search?q=ENGG2013+Unit+24&oq=ENGG2013+Unit+24&aqs=chrome..69i57.2731j0j7&sourceid=chrome&ie=UTF-8

https://www.google.com.bd/search?q=ENGG2013+Unit+24&oq=ENGG2013+Unit+24&aqs=chrome..69i57.2731j0j7&sourceid=chrome&ie=UTF-8#q=minggu-2-1-engineering-mathematics-differential-equations

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THANKS all.

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Any Question?